ITPEC FE Morning April 2023 Question 18
ITPEC FE Morning April 2023 — Question 18 of 80
SRTF Average Waiting Time — calculate avg waiting time under preemptive Shortest Remaining Time First scheduling.
Given processes:
| Process | Arrival | Burst |
|---|---|---|
| P1 | 0 | 8 |
| P2 | 2 | 3 |
| P3 | 3 | 7 |
| P4 | 5 | 4 |
Step 1: Build the timeline
- •
t=0: P1 starts (only process available, remaining=8) - •
t=2: P2 arrives (burst=3). P1 remaining=6.3 < 6→ preempt P1, run P2 - •
t=3: P3 arrives (burst=7). P2 remaining=2.2 < 7→ P2 continues - •
t=5: P2 completes. P4 arrives (burst=4). Remaining: P1=6, P3=7, P4=4. Min=4 → run P4 - •
t=9: P4 completes. Remaining: P1=6, P3=7. Min=6 → run P1 - •
t=15: P1 completes. Run P3 - •
t=22: P3 completes
Step 2: Calculate waiting times
- •
Waiting Time = (Completion - Arrival) - Burst - •P1:
(15 - 0) - 8= 7 - •P2:
(5 - 2) - 3= 0 - •P3:
(22 - 3) - 7= 12 - •P4:
(9 - 5) - 4= 0
Result: (7 + 0 + 12 + 0) / 4 = 4.75
Why not others:
- (a) 3.0 — underestimates P1 and P3 waiting
- (c) 5.5 — likely a non-preemptive SJF calculation error
- (d) 10.25 — sum of waiting times (19), not the average
Key rule: In SRTF, always draw a timeline checking remaining times at each arrival and completion event. Use Waiting = Turnaround - Burst for each process.
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